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Gilda walks to the train station. If she walks at the rate of 3 mph, she misses her train by 7 minutes. However, if she walks at the rate of 4 mph, she reaches the station 5 minutes before the arrival of the train. Find the distance Gilda walks to the station.

David knew he made a mistake when he calculated that Gilda walks 123 miles to the station. Read through David's calculations:Using d = rt, the distance is the same, but the rate and time are different.If Gilda misses the train, it means the time t needs 7 more minutes so d = 3(t + 7).If she gets to the station 5 minutes early means the time t can be 5 minutes less so d = 4(t - 5).3(t + 7) = 4(t - 5)3t + 21 = 4t - 20t = 41d = rt, so d = 3(41) = 123Find David's mistake in his calculations. In two or more complete sentences, explain his mistake. Include the correct calculations and solutions in your answer.

Sagot :

Ananda26idn
We'll represent the distance as d and the time as t. 
d= 3(t+7)            because Gilda is 7 minutes late when travelling 3 mph
d= 4(t-5)             because Gilda is 5 minutes early at 4 mph

d= 3t + 21
d= 4t -20
3t +21 =4t -20
41=t
d= 3(41) +21= 144
The distance is 144 miles. 
Kate200468idn
[tex]d=3(t + 7)\ \ \ if\ Gilda\ walks\ of\ 3mph\\.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ she\ is\ 7\ minutes\ after\ departure\ of\ the\ train\\\\ d= 4(t - 5)\ \ \ if\ Gilda\ walks\ of\ 4mph\ \\.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ she\ is\ 5\ minutes\ before\ the\ train\ \\\\3(t+7)=4(t-5)\\\\3t + 21 = 4t - 20 \\\\t = 41\ \ \ \Rightarrow\ \ \ d=3(t+7)=3\cdot(41+7)=3\cdot48=144\\\\Ans.\ The\ distance\ to\ the\ station\ is\ 144\ miles.[/tex]
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